A 1.85 mass % aqueous solution urea (CO (NH2) 2) has a density of 1.05 g/mL. Calculate the molality of the solution. Give your answer to 2 decimal places.

Molality for this solution is 0.31 m

Explanation:

Molality → Mol of solute / kg of solvent

Mass of solute = 1.85 g (CO (NH₂) ₂

% by mass, means mass of solute in 100 g of solution

Mass of solution = Mass of solute + Mass of solvent

As 1.85 g is the mass of solute, the mass of solvent would be 98.15 g to complete the 100g of solution.

Let's convert the mass of solvent (g to kg)

1 g = 1*10⁻³ kg

98.2 g. 1*10⁻³ kg / 1g = 0.0982 kg

Let's convert the mass of solute to moles (mass / molar mass)

1.85 g / 60 g/mol = 0.0308 mol

Molality = 0.0308 mol / 0.0982 kg → 0.31 m